Question: Solve for $x$ : $ 7|x + 5| + 1 = -3|x + 5| + 4 $
Add $ {3|x + 5|} $ to both sides: $ \begin{eqnarray} 7|x + 5| + 1 &=& -3|x + 5| + 4 \\ \\ { + 3|x + 5|} && { + 3|x + 5|} \\ \\ 10|x + 5| + 1 &=& 4 \end{eqnarray} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} 10|x + 5| + 1 &=& 4 \\ \\ { - 1} &=& { - 1} \\ \\ 10|x + 5| &=& 3 \end{eqnarray} $ Divide both sides by ${10}$ $ \dfrac{10|x + 5|} {{10}} = \dfrac{3} {{10}} $ Simplify: $ |x + 5| = \dfrac{3}{10}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 5 = -\dfrac{3}{10} $ or $ x + 5 = \dfrac{3}{10} $ Solve for the solution where $x + 5$ is negative: $ x + 5 = -\dfrac{3}{10} $ Subtract ${5}$ from both sides: $ \begin{eqnarray} x + 5 &=& -\dfrac{3}{10} \\ \\ {- 5} && {- 5} \\ \\ x &=& -\dfrac{3}{10} - 5 \end{eqnarray} $ Change the ${ - 5}$ to an equivalent fraction with a denominator of $10$ $ x = - \dfrac{3}{10} {- \dfrac{50}{10}} $ $ x = -\dfrac{53}{10} $ Then calculate the solution where $x + 5$ is positive: $ x + 5 = \dfrac{3}{10} $ Subtract ${5}$ from both sides: $ \begin{eqnarray} x + 5 &=& \dfrac{3}{10} \\ \\ {- 5} && {- 5} \\ \\ x &=& \dfrac{3}{10} - 5 \end{eqnarray} $ Change the ${ - 5}$ to an equivalent fraction with a denominator of $10$ $ x = \dfrac{3}{10} {- \dfrac{50}{10}} $ $ x = -\dfrac{47}{10} $ Thus, the correct answer is $x = -\dfrac{53}{10} $ or $x = -\dfrac{47}{10} $.